Question

When \(30\,\text{mL}\) of \(0.2\,\text{M}\) \(NH_4OH\) is added to \(30\,\text{mL}\) of \(2\,\text{M}\) \(NH_4Cl\) solution. If the pH of the buffer formed is \(8.2\), what is the pKb of \(NH_4OH\)?

• A. $7.2$
• B. $5.8$
• C. $6.8$
• D. $4.8$

Hint:

For equal volumes of mixture components, you can use the direct millimoles instead of recalculating the final molarity of the salt and base. This saves valuable calculation time during exams.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This numerical problem asks to calculate the dissociation constant exponent ($\text{pK}_b$) of a weak base ($\text{NH}_4\text{OH}$) in a basic buffer solution of known pH and concentration.

Step 2: Key Formula or Approach:
For a basic buffer solution (weak base + conjugate salt), we use the Henderson-Hasselbalch equation:
\[ \text{pOH} = \text{pK}_b + \log\left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \] Also, the relationship between pH and pOH at $298\text{ K}$ is:
\[ \text{pH} + \text{pOH} = 14 \implies \text{pOH} = 14 - \text{pH} \]

Step 3: Detailed Explanation:

• First, let us calculate the pOH of the buffer solution from the given pH:
\[ \text{pOH} = 14 - \text{pH} \] \[ \text{pOH} = 14 - 8.2 = 5.8 \]
• Next, we calculate the number of millimoles of the weak base ($\text{NH}_4\text{OH}$) and the salt ($\text{NH}_4\text{Cl}$) in the mixture:
- Millimoles of weak base ($\text{NH}_4\text{OH}$):
\[ \text{Volume} \times \text{Molarity} = 30\text{ ml} \times 0.2\text{ M} = 6\text{ mmol} \] - Millimoles of salt ($\text{NH}_4\text{Cl}$):
\[ \text{Volume} \times \text{Molarity} = 30\text{ ml} \times 2\text{ M} = 60\text{ mmol} \]
• Since both components are present in the same total volume of $60\text{ ml}$, the ratio of their molar concentrations is equal to the ratio of their millimoles:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = \frac{60\text{ mmol}}{6\text{ mmol}} = 10 \]
• Now, we substitute these values into the Henderson-Hasselbalch equation:
\[ 5.8 = \text{pK}_b + \log(10) \]
• We know that $\log_{10}(10) = 1$. Therefore:
\[ 5.8 = \text{pK}_b + 1 \] \[ \text{pK}_b = 5.8 - 1 \] \[ \text{pK}_b = 4.8 \]
• Thus, the $\text{pK}_b$ of ammonium hydroxide is $4.8$.

Step 4: Final Answer:
The $\text{pK}_b$ of $\text{NH}_4\text{OH}$ is $4.8$.